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WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Geometric Transformations and WallpaperGroups
Isometries of the Plane
Lance Drager
Department of Mathematics and StatisticsTexas Tech University
Lubbock, Texas
2010 Math Camp
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Outline
1 Introduction
2 Orthogonal MatricesOrthogonal Matrices and Dot ProductClassifying Orthogonal MatricesExercises
3 Classifying IsometriesTranslationsFirst ClassificationFinal ClassificationExercises
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Isometries
• A transformation T of the plane is an isometry if it isone-to-one and onto and
dist(T (p),T (q)) = dist(p, q), for all points p and q.
• If S and T are isometries, so is ST , where(ST )(p) = S(T (p)).
• If T is an isometry, so is T−1.
• Our goal is to find all the isometries.
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Orthogonal Matrices
• A matrix A is orthogonal if ‖Ax‖ = ‖x‖ for all x ∈ R2.
• An orthogonal matrix gives an isometry of the plane since
dist(Ap,Aq) = ‖Ap − Aq‖= ‖A(p − q)‖ = ‖p − q‖ = dist(p, q)
• If A and B are orthogonal, so is AB.
• If A is orthogonal, it is invertible and A−1 is orthogonal.
• Rotation matrices are orthogonal.
y
x
‖p− q‖
p
q
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Outline
1 Introduction
2 Orthogonal MatricesOrthogonal Matrices and Dot ProductClassifying Orthogonal MatricesExercises
3 Classifying IsometriesTranslationsFirst ClassificationFinal ClassificationExercises
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Dot Product 1
TheoremA is orthogonal if and only if Ax · Ay = x · y for all x , y ∈ R2.Thus, an orthogonal matrix preserves angles.
Proof.(=⇒)
‖Ax‖2 = Ax · Ax = x · x = ‖x‖2
(⇐=) Recall
2x · y = ‖x‖2 + ‖y‖2 − ‖x − y‖2.
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Dot Product 2
Proof Continued.So, we have
2Ax · Ay = ‖Ax‖2 + ‖Ay‖2 − ‖Ax − Ay‖2
= ‖Ax‖2 + ‖Ay‖2 − ‖A(x − y)‖2 (distributive law)
= ‖x‖2 + ‖y‖2 − ‖x − y‖2 (A is orthogonal)
= 2x · y ,
and dividing by 2 gives the result.
TheoremA matrix is orthogonal if and only if its columns are orthogonalunit vectors.
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Dot Product 3
Proof.(=⇒)
‖Aei‖ = ‖ei‖ = 1.
Ae1 · Ae2 = e1 · e2 = 0.
(⇐=) Let A = [u | v ] where u and v are orthogonal unitvectors. Note Ax = x1u + x2v .
‖Ax‖2 = Ax · Ax
= (x1u + x2v) · (x1u + x2v)
= x21 u · u + 2x1x2u · v + x2
2 v · v= x2
1 (1) + 2x1x2(0) + x22 (1)
= x21 + x2
2 = ‖x‖2
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Outline
1 Introduction
2 Orthogonal MatricesOrthogonal Matrices and Dot ProductClassifying Orthogonal MatricesExercises
3 Classifying IsometriesTranslationsFirst ClassificationFinal ClassificationExercises
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Classifying Orthogonal Matrices 1
• If A is orthogonal, Col1(A) = (cos(θ), sin(θ)) for some θ.So the possibilities are
A =
[cos(θ) − sin(θ)sin(θ) cos(θ)
]= R(θ),
A =
[cos(θ) sin(θ)sin(θ) − cos(θ)
]= S(θ).
• In the first case A = R(θ) is a rotation.
• What about S(θ)?
• There are orthogonal unit vectors u and v so thatS(θ)u = u and S(θ)v = −v .
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Classifying Orthogonal Matrices 1
• If A is orthogonal, Col1(A) = (cos(θ), sin(θ)) for some θ.So the possibilities are
A =
[cos(θ) − sin(θ)sin(θ) cos(θ)
]= R(θ),
A =
[cos(θ) sin(θ)sin(θ) − cos(θ)
]= S(θ).
• In the first case A = R(θ) is a rotation.
• What about S(θ)?
• There are orthogonal unit vectors u and v so thatS(θ)u = u and S(θ)v = −v .
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Classifying Orthogonal Matrices 2
• In fact, let u = (cos(θ/2), sin(θ/2)) andv = (− sin(θ/2), cos(θ/2)).
•
S(θ)u =
[cos(θ) sin(θ)sin(θ) − cos(θ)
] [cos(θ/2)sin(θ/2)
]=
[cos(θ) cos(θ/2) + sin(θ) sin(θ/2)sin(θ) cos(θ/2)− cos(θ) sin(θ/2)
]
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Classifying Orthogonal Matrices 2
• In fact, let u = (cos(θ/2), sin(θ/2)) andv = (− sin(θ/2), cos(θ/2)).
•
S(θ)u =
[cos(θ) sin(θ)sin(θ) − cos(θ)
] [cos(θ/2)sin(θ/2)
]=
[cos(θ) cos(θ/2) + sin(θ) sin(θ/2)sin(θ) cos(θ/2)− cos(θ) sin(θ/2)
]=
[cos(θ − θ/2)sin(θ − θ/2)
]=
[cos(θ/2)sin(θ/2)
]= u.
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Classifying Orthogonal Matrices 3
•
S(θ)v =
[cos(θ) sin(θ)sin(θ) − cos(θ)
] [− sin(θ/2)
cos(θ/2)
]=
[− cos(θ) sin(θ/2) + sin(θ) cos(θ/2)− sin(θ) sin(θ/2)− cos(θ) cos(θ/2)
]=
[sin(θ − θ/2)
− cos(θ − θ/2)
]=
[sin(θ/2)
− cos(θ/2)
]= −
[− sin(θ/2)
cos(θ/2)
]= −v .
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Classifying Orthogonal Matrices 4
• S(θ)u = u and S(θ)v = −v . If x = tu + sv , thenS(θ)x = tu − sv .
x
y
u
tu
v
sv
−sv
M
x
S(θ)x
• S(θ) is a reflection with its mirror line at an angle of θ/2.
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Classifying Orthogonal Matrices 4
• S(θ)u = u and S(θ)v = −v . If x = tu + sv , thenS(θ)x = tu − sv .
x
y
u
tu
v
sv
−sv
M
x
S(θ)x
• S(θ) is a reflection with its mirror line at an angle of θ/2.
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Outline
1 Introduction
2 Orthogonal MatricesOrthogonal Matrices and Dot ProductClassifying Orthogonal MatricesExercises
3 Classifying IsometriesTranslationsFirst ClassificationFinal ClassificationExercises
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Orthogonal Matix Exercises 1
Exercises
• S(θ)2 = I , so S(θ)−1 = S(θ).
• Let A be an orthogonal matrix. Then A is a rotation (orthe identity) if and only if det(A) = 1 and A is a reflectionif and only if det(A) = −1.
• If A =
[a cb d
]define the transpose of A by
AT =
[a bc d
]. Show that A is orthogonal if and only if
A−1 = AT .
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Exercises on Orthogonal Matrices2
Exercises Continued
• Use the addition laws for sine and cosine to verify theimportant identities
R(θ)S(φ) = S(θ + φ),
S(φ)R(θ) = S(φ− θ),
S(θ)S(φ) = R(θ − φ)
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Outline
1 Introduction
2 Orthogonal MatricesOrthogonal Matrices and Dot ProductClassifying Orthogonal MatricesExercises
3 Classifying IsometriesTranslationsFirst ClassificationFinal ClassificationExercises
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Translations
• For v ∈ R2, define Tv : R2 → R2 by Tv (x) = x + v . Wesay Tv is translation by v.
• TuTv = Tu+v and T−1v = T−v .
• Translations are isometries
dist(Tv (x),Tv (y)) = ‖Tv (x)− Tv (y)‖= ‖(x + v)− (y + v)‖= ‖x + v − y − v‖= ‖x − y‖= dist(x , y).
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Outline
1 Introduction
2 Orthogonal MatricesOrthogonal Matrices and Dot ProductClassifying Orthogonal MatricesExercises
3 Classifying IsometriesTranslationsFirst ClassificationFinal ClassificationExercises
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Three Distances Determine a Point
• Let p1, p2 and p3 be noncolinear points. A point x isuniquely determined by the three numbers r1 = dist(x , p1),r2 = dist(x , p2) and r3 = dist(x , p3).
p1 p2
p3
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
First Classification
• If an isometry T fixes three noncolinear points p1, p2, p3
then T is the identity.
dist(x , pi ) = dist(T (x),T (pi )) = dist(T (x), pi ), i = 1, 2, 3,
=⇒ x = T (x).
TheoremEvery isometry T can be written as T (x) = Ax + v where A isan orthogonal matrix, i.e., T is multiplication by an orthogonalmatrix followed by a translation.
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Proof of First Classification
Proof.
1 The points 0, e1, e2 are noncolinear.
2 Let u = −T (0). Then TuT (0) = 0.
3 dist(TuT (e1), 0) = 1, so we can find a rotation matrix Rso that RTuT (e1) = e1.
4 RTuT (e2) = ±e2.
5 If RTuT (e2) = −e2, let S = S(0) be reflection throughthe x-axis, otherwise let S = I .
6 SRTu(0) = 0, SRTuT (e1) = e2, and SRTuT (e2) = e2
7 SRTuT = I , so T = T−uR−1S−1
8 Let A = R−1S−1 and v = −u. Then T (x) = Ax + v
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Proof of First Classification
Proof.
1 The points 0, e1, e2 are noncolinear.
2 Let u = −T (0). Then TuT (0) = 0.
3 dist(TuT (e1), 0) = 1, so we can find a rotation matrix Rso that RTuT (e1) = e1.
4 RTuT (e2) = ±e2.
5 If RTuT (e2) = −e2, let S = S(0) be reflection throughthe x-axis, otherwise let S = I .
6 SRTu(0) = 0, SRTuT (e1) = e2, and SRTuT (e2) = e2
7 SRTuT = I , so T = T−uR−1S−1
8 Let A = R−1S−1 and v = −u. Then T (x) = Ax + v
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Proof of First Classification
Proof.
1 The points 0, e1, e2 are noncolinear.
2 Let u = −T (0). Then TuT (0) = 0.
3 dist(TuT (e1), 0) = 1, so we can find a rotation matrix Rso that RTuT (e1) = e1.
4 RTuT (e2) = ±e2.
5 If RTuT (e2) = −e2, let S = S(0) be reflection throughthe x-axis, otherwise let S = I .
6 SRTu(0) = 0, SRTuT (e1) = e2, and SRTuT (e2) = e2
7 SRTuT = I , so T = T−uR−1S−1
8 Let A = R−1S−1 and v = −u. Then T (x) = Ax + v
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Proof of First Classification
Proof.
1 The points 0, e1, e2 are noncolinear.
2 Let u = −T (0). Then TuT (0) = 0.
3 dist(TuT (e1), 0) = 1, so we can find a rotation matrix Rso that RTuT (e1) = e1.
4 RTuT (e2) = ±e2.
5 If RTuT (e2) = −e2, let S = S(0) be reflection throughthe x-axis, otherwise let S = I .
6 SRTu(0) = 0, SRTuT (e1) = e2, and SRTuT (e2) = e2
7 SRTuT = I , so T = T−uR−1S−1
8 Let A = R−1S−1 and v = −u. Then T (x) = Ax + v
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Proof of First Classification
Proof.
1 The points 0, e1, e2 are noncolinear.
2 Let u = −T (0). Then TuT (0) = 0.
3 dist(TuT (e1), 0) = 1, so we can find a rotation matrix Rso that RTuT (e1) = e1.
4 RTuT (e2) = ±e2.
5 If RTuT (e2) = −e2, let S = S(0) be reflection throughthe x-axis, otherwise let S = I .
6 SRTu(0) = 0, SRTuT (e1) = e2, and SRTuT (e2) = e2
7 SRTuT = I , so T = T−uR−1S−1
8 Let A = R−1S−1 and v = −u. Then T (x) = Ax + v
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Proof of First Classification
Proof.
1 The points 0, e1, e2 are noncolinear.
2 Let u = −T (0). Then TuT (0) = 0.
3 dist(TuT (e1), 0) = 1, so we can find a rotation matrix Rso that RTuT (e1) = e1.
4 RTuT (e2) = ±e2.
5 If RTuT (e2) = −e2, let S = S(0) be reflection throughthe x-axis, otherwise let S = I .
6 SRTu(0) = 0, SRTuT (e1) = e2, and SRTuT (e2) = e2
7 SRTuT = I , so T = T−uR−1S−1
8 Let A = R−1S−1 and v = −u. Then T (x) = Ax + v
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Proof of First Classification
Proof.
1 The points 0, e1, e2 are noncolinear.
2 Let u = −T (0). Then TuT (0) = 0.
3 dist(TuT (e1), 0) = 1, so we can find a rotation matrix Rso that RTuT (e1) = e1.
4 RTuT (e2) = ±e2.
5 If RTuT (e2) = −e2, let S = S(0) be reflection throughthe x-axis, otherwise let S = I .
6 SRTu(0) = 0, SRTuT (e1) = e2, and SRTuT (e2) = e2
7 SRTuT = I , so T = T−uR−1S−1
8 Let A = R−1S−1 and v = −u. Then T (x) = Ax + v
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Proof of First Classification
Proof.
1 The points 0, e1, e2 are noncolinear.
2 Let u = −T (0). Then TuT (0) = 0.
3 dist(TuT (e1), 0) = 1, so we can find a rotation matrix Rso that RTuT (e1) = e1.
4 RTuT (e2) = ±e2.
5 If RTuT (e2) = −e2, let S = S(0) be reflection throughthe x-axis, otherwise let S = I .
6 SRTu(0) = 0, SRTuT (e1) = e2, and SRTuT (e2) = e2
7 SRTuT = I , so T = T−uR−1S−1
8 Let A = R−1S−1 and v = −u. Then T (x) = Ax + v
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Outline
1 Introduction
2 Orthogonal MatricesOrthogonal Matrices and Dot ProductClassifying Orthogonal MatricesExercises
3 Classifying IsometriesTranslationsFirst ClassificationFinal ClassificationExercises
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Notation and Algebra
• If T (x) = Ax + v write T = (A | v).
• Calculate the product (composition) in this notation
(A | u)(B | v)x = (A | u)(Bx + v)
= A(Bx + v) + u
= ABx + Av + u
= (AB | u + Av)x .
• (A | u)(B | v) = (AB | u + Av)
• The identity transformation is (I | 0). Translation by v is(I | v).
• (A | u)−1 = (A−1 | − A−1u)
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Rotations
• Consider (R | v) where R = R(θ) 6= I is a rotation.
• Look for a fixed point p, i.e., (R | v)p = p
Rp + v = p
⇐⇒ p − Rp = v
⇐⇒ (I − R)p = v .
• If (I − R) is invertible, there is a unique solution p.
• (I − R) is invertible because(I − R)x = 0 ⇐⇒ x = Rx ⇐⇒ x = 0. (Use the BigTheorem.)
• There is a unique fixed point p, and we can write(R | v) = (R | p − Rp).
• (R | p − Rp)x = Rx + p − Rp = p + R(x − p).
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Picture of Rotation
• y = (R | p − Rp)x = p + R(x − p) says that this isometryis rotation though angle θ around the point p,which iscalled the center of rotation or rotocenter.
x
y
p
xx− pθ
R(x− p)
y
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Isometries with a Reflection Matrix
• Consider (S |w) where S = S(θ) is a reflection matrix.
• Let u and v be the vectors with Su = u and Sv = −v .We can write w = αu + βv , so (S |w) = (S |αu + βv).
• First Case: α = 0. So we have (S |βv).
• The point p = βv/2 is fixed.
(S |βv)p = S(βv/2) + βv = −βv/2 + βv = βv/2 = p.
• The fixed points are exactly x = p + tu.
(S | 2p)(p + tu) = Sp + tSu + 2p = −p + tu + 2p = p + tu.
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Isometries with a Reflection Matrix
• Consider (S |w) where S = S(θ) is a reflection matrix.
• Let u and v be the vectors with Su = u and Sv = −v .We can write w = αu + βv , so (S |w) = (S |αu + βv).
• First Case: α = 0. So we have (S |βv).
• The point p = βv/2 is fixed.
(S |βv)p = S(βv/2) + βv = −βv/2 + βv = βv/2 = p.
• The fixed points are exactly x = p + tu.
(S | 2p)(p + tu) = Sp + tSu + 2p = −p + tu + 2p = p + tu.
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Isometries with a Reflection Matrix
• Consider (S |w) where S = S(θ) is a reflection matrix.
• Let u and v be the vectors with Su = u and Sv = −v .We can write w = αu + βv , so (S |w) = (S |αu + βv).
• First Case: α = 0. So we have (S |βv).
• The point p = βv/2 is fixed.
(S |βv)p = S(βv/2) + βv = −βv/2 + βv = βv/2 = p.
• The fixed points are exactly x = p + tu.
(S | 2p)(p + tu) = Sp + tSu + 2p = −p + tu + 2p = p + tu.
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
The Line of Fixed Points
• The line through p parallel to u is parametrized byx = p + tu, for t ∈ R.
x
y
x
p
tu
v u
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Reflection Picture
• We can write a point x as x = p + tu + sv . Theny = (S | 2p)x = p + tu − sv .
x
y
M
p
x
y
tu
sv
−sv
tu + sv
tu− sv
v u
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Reflections and Glides
• (S |βv) = (S | 2p) is reflection through the mirror line Mgiven by y = p + tu.
• Second Case: α 6= 0. We have (S |αu + βv).
(S |αu + βv) = (I |αu)(S |βv)
This is a reflection followed by a translation parallel to themirror line. This is called a glide refection or just a glide.The mirror line of the reflection is called the glide line.
• A glide has no fixed points.
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Glide Reflection Picture
• A glide with a horizontal glide line, and the translationvector shown in blue.
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Classification Theorem
TheoremEvery isometry of the plane falls into on of the following fivemutually exclusive classes.
1 The identity.
2 A translation (not the identity).
3 A rotation about some point (not the identity);
4 A reflection through some mirror line.
5 A glide along some glide line.
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Outline
1 Introduction
2 Orthogonal MatricesOrthogonal Matrices and Dot ProductClassifying Orthogonal MatricesExercises
3 Classifying IsometriesTranslationsFirst ClassificationFinal ClassificationExercises
WallpaperGroups
Lance Drager
Introduction
OrthogonalMatrices
OrthogonalMatrices andDot Product
ClassifyingOrthogonalMatrices
Exercises
ClassifyingIsometries
Translations
FirstClassification
FinalClassification
Exercises
Exercises on Isometries
Exercises
1 Consider the cases for the product T1T2 of two isometriesT1 and T2. Describe what happens geometrically (e.g.,rotation angle, location of the mirror line, etc.). In whatcases do the isometries commute, i.e., when doesT1T2 = T2T1
2 Project: For a rotation (R | v) give a geometric descriptionof how to find the rotocenter p = (I − R)−1v .
3 As part of the first exercise, given two rotations (withpossibly different rotocenters) show that the compositionis a rotation or a translation.
4 Project: Given two rotations with different rotocentersgive a geometric description of how to find the rotocenterof the composition.